Let $f, g$ be newforms of level $N_f, N_g$ (possibly different) with $N_f, N_g |N$, and assume that they have same eigenvalue for Hecke operators $T_p$ with $(p, N) = 1$. Then $N_f = N_g$ and $g = cf$ for some constant $c$.
My friend gave one proof using the Galois representation attached to these forms and Carayol's theorem on the conductor of these representations, but this seems a little overkill for me. I wonder if there's a simpler proof for this. (It seems that there's a proof in Miyake's book using $L$-function).
Here's the proof. We will assume strong multiplicity one theorem for new forms of the same level. So the only thing we need is to prove $N_{f} = N_{g}$. Let $\rho_{f, \ell}, \rho_{g, \ell}$ be the attached $\ell$-adic Galois representations. Since $\mathrm{tr}(\rho_{f, \ell}(\mathrm{Frob}_{p})) = a_{p}(f) = a_p(g) = \mathrm{tr}(\rho_{g, \ell}(\mathrm{Frob}_{p}))$ for all $p\nmid \ell N$, we have $\rho_{f, \ell} \simeq \rho_{g, \ell}$. Carayol's theorem states that the $\ell$-free part of $\mathrm{cond}(\rho_{f, \ell})$ and $N_{f}$ are the same, so by choosing different $\ell$'s, we get $N_{f} = N_{g}$.